Question

For the cell $ Ag_{(s)}| Ag^+_{(aq)} | | Cu^{2+}_{(aq)} | Cu_{(s)} $ , the reduction potentials of the left and right hand electrodes are $ 0.337 $ and $ 0.799 $ volts, the cell $e.m.f$ is

• A. $ -1.136 $ volt
• B. $ 1.136 $ volt
• C. $ - 0.462 $ volt
• D. $ 0.462 $ volt

Answer 1

Answer: (D)

$E^{\circ}_{\text{cell}} = E^{\circ} _{\text{(right hand electrode)-}}$
$Ag(s) | Ag^+(aq)| | Cu^{2+} (aq)| Cu\, E^{\circ} _{\text{(Left hand electrode)}}$
$E^{\circ} /V$
$LHS \,\,\, RHS$
Electrode
$Ag^+ +2e^- \rightleftharpoons Ag \,\,\,0.337$
$ Cu^{2+} + 2e^- \rightleftharpoons Cu \,\,\,0.799$
$E^{\circ}_{\text{cell}} = RHS - LHS$
$ = 0.799 - 0.337 = 0.462$ Volt